A) \[\pm 7\]
B) \[\pm 5\]
C) \[\pm 10\]
D) \[\pm 9\]
Correct Answer: C
Solution :
Given, line is \[3x+y+k=0\] \[\Rightarrow \] \[y=-3x-k\] And equation of circle is \[{{x}^{2}}+{{y}^{2}}=10\] Here, \[{{a}^{2}}=10,\] \[m=-3,\] \[c=-k\] If given line touches the circle, then Length of intercept \[=0\] \[\Rightarrow \] \[2\sqrt{\frac{{{a}^{2}}(1+{{m}^{2}})-{{c}^{2}}}{1+{{m}^{2}}}}=0\] \[\Rightarrow \] \[2\sqrt{\frac{10(1+9)-{{k}^{2}}}{1+9}}=0\] \[\Rightarrow \] \[\sqrt{100-{{k}^{2}}}=0\] \[\Rightarrow \] \[100-{{k}^{2}}=0\] \[\Rightarrow \] \[k=\pm 10\] Alternative: If the given line is tangent to the circle, then the length of the perpendicular from the centre upon the line is equal to the radius of the circle. ie, \[\left| \frac{a{{x}_{1}}+b{{y}_{1}}+c}{\sqrt{{{a}^{2}}+{{b}^{2}}}} \right|=r\] \[\Rightarrow \] \[\left| \frac{3\times 0+6\times 0+k}{\sqrt{{{(3)}^{2}}+{{(1)}^{2}}}} \right|=\sqrt{10}\] \[\Rightarrow \] \[\left| \frac{k}{\sqrt{10}} \right|=\sqrt{10}\] \[\Rightarrow \] \[k=\sqrt{100}\] \[\Rightarrow \] \[k=\pm 10\]
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